College of the Holy Cross, Spring 2008 Math 132, Solutions for Midterm Exam 1 (All Sections) Wednesday, February 20

I. (10) State the definition of the definite integral ∫_{a}^{b} f(x) dx.

Solution: If f(x) is continuous on [a, b] and Δx = \frac{b−a}{n}, then the definite integral is the limit of the Riemann sums

∫_{a}^{b} f(x) dx = lim_{n→∞} ∑_{i=1}^{n} f(x_{i}^{∗}) Δx

for all choices of x_{i}^{∗} ∈ [x_{i−1}, x_{i}].

II. Compute the derivatives of each of the following functions defined by integrals.

A. (5) f(x) = ∫_{1}^{x} e^{t^{2}} dt

Solution: By the first part of the Fundamental Theorem of Calculus,

f′(x) = e^{x^{2}}.

B. (5) g(x) = ∫_{x^{2}}^{4} \frac{cos(t)}{t^{2}} dt

Solution: Reversing the limits of integration,

g(x) = − ∫_{4}^{x^{2}} \frac{cos(t)}{t^{2}} dt.

Then by the first part of the Fundamental Theorem and the Chain Rule,

g′(x) = −\frac{cos(x^{2})}{x^{4}} · 2x = −\frac{2 cos(x^{2})}{x^{3}}.

III. The following graph (made up of straight line segments) shows y = f(t) for 0 ≤ t ≤ 4.

Given: f(1) = 2, f(2) = −2, f(3) = −1, and f(4) = 0, The function F is defined by F (x) = ∫_{0}^{x} f(t) dt.

A. (5) Determine the values F (x) for x = 0, 1, 2, 3, 4 and enter them in the following table.

Solution: By computing areas of trapezoids, we have:

\begin{matrix}x & 0 & 1 & 2 & 3 & 4 \\ F (x) & 0 & 3/2 & 3/2 & 0 & -1/2 \end{matrix}

B. (5) Does F (x) have any critical points? If so, say where. If not say why not.

Solution: By the first part of the Fundamental Theorem, F′(x) = f(x) = 0 when x = 3/2 and x = 4. (Since x = 4 is an endpoint of the interval, we will also accept just x = 3/2.)

C. (5) Over which interval(s) is F (x) concave down?

Solution: F (x) is concave down on intervals where F′′(x) < 0, or F′(x) = f(x) is decreasing. This occurs for 1 < x < 2 here.

IV.

A. (5) Integrate with a suitable u-substitution: ∫_{0}^{1}(4x^{3} + 1)^{3/5}x^{2} dx.

Solution: Note that the x^{2} outside the power is du up to a constant for u = 4x^{3} + 1. So do this substitution where du = 12x^{2} dx. So

∫_{0}^{1}(4x^{3} + 1)^{3/5}x^{2} dx = \frac{1}{12} ∫_{u=1}^{u=5} u^{3/5} du
= \frac{5}{96}u^{8/5}|_{u=1}^{u=5}
= \frac{5}{96} (5^{8/5} − 1)

B. (5) Integrate with a suitable u-substitution: ∫ \frac{x sin(3x^{2})}{cos(3x^{2}) + 1} dx.

Solution: Let u = cos(3x^{2}) + 1. Then du = − sin(3x^{2})(6x) dx by the Chain Rule. So

∫ \frac{x sin(3x^{2})}{cos(3x^{2}) + 1} dx = −\frac{1}{6} ∫ \frac{du}{u}
= −\frac{1}{6} ln |u| + C
= −\frac{1}{6} ln | cos(3x^{2}) + 1| + C.

C. (7.5) Integrate by parts: ∫ x^{2} sin(5x) dx

Solution: We let u = x^{2}, dv = sin(5x) dx, so du = 2x dx and v = −\frac{1}{5} cos(5x). Applying the parts formula,

∫ x^{2} sin(5x) dx = −\frac{x^{2}}{5} cos(5x) + \frac{2}{5} ∫ x cos(5x) dx.

Now do parts again on the remaining integral with u = x, dv = cos(5x) dx, so du = dx and v = \frac{1}{5} sin(5x). This gives:

∫ x^{2} sin(5x) dx = −\frac{x^{2}}{5} cos(5x) + \frac{2}{5} (\frac{x}{5} sin(5x) − \frac{1}{5} ∫ sin(5x) dx)
= −\frac{x^{2}}{5} cos(5x) + \frac{2x}{25} sin(5x) + \frac{2}{125} cos(5x) + C

D. (7.5) Integrate with the partial fraction method: ∫ \frac{3x^{2} + 1}{x^{2} + 5x + 6} dx

Solution: Since the degree of the top is equal to the degree of the bottom, we divide first: 3x^{2} + 1 = 3(x^{2} + 5x + 6) − 15x − 17 so

∫ \frac{3x^{2} + 1}{x^{2} + 5x + 6} dx = ∫ 3 − \frac{15x + 17}{x^{2} + 5x + 6} dx.

Now x^{2} + 5x + 6 = (x + 2)(x + 3) so the partial fractions are

\frac{15x + 17}{(x + 2)(x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3}.

Clearing denominators,

15x + 17 = A(x + 3) + B(x + 2),

so with x = −2 we see A = −13 and with x = −3 we see B = 28. Then

∫ 3− \frac{15x + 17}{x^{2} + 5x + 6} dx = ∫ 3−(\frac{−13}{x + 2} + \frac{28}{x + 3}) dx = 3x+13 ln |x+2|−28 ln |x+3|+C.

E. (10) Integrate via trigonometric substitution: ∫ \sqrt{36 − x^{2}} dx Solution: From the form of the integrand, we use x = 6 sin θ, so dx = 6 cos θ dθ and the integral becomes

∫ \sqrt{36 − x^{2}} dx = ∫ \sqrt{36(1 − sin^{2} θ)} 6 cos θ dθ = 36 ∫ cos^{2} θ dθ.

To integrate this, we use the half-angle formula:

36 ∫ cos^{2} θ dθ = 36 ∫ \frac{1}{2}(1 + cos(2θ)) dθ = 18θ + 9 sin(2θ) + C

Now we convert back to expressions in the original variable x. From the substitution x = 6 sin θ, we have θ = sin^{−1} (\frac{x}{6}) and sin(2θ) = 2 sin(θ) cos(θ) = 2\frac{x}{6} \frac{\sqrt{36−x^{2}}}{6} = \frac{x\sqrt{36−x^{2}}}{18} . The integral equals

18 sin^{−1} (\frac{x}{6}) + \frac{x\sqrt{36 − x^{2}}}{2} + C.

(Note that you can check your work using #39 in the table!)

V. Integrate using any applicable method or the table. If you do use the table, give the number of the entry you are using.

A. (10) ∫ tan x sec^{4} x dx.

Solution: Method 1: Since the derivative of u = sec x is du = sec x tan x, the form here is

∫ sec^{3} x(sec x tan x) dx = ∫ u^{3} du = \frac{1}{4}u^{4} + C = \frac{1}{4} sec^{4} x + C.

Method 2: Since the derivative of tan x is sec^{2} x, this one can also be done by splitting off two powers of sec x and converting the other sec^{2} x to powers of tan x:

∫ tan x sec^{4} x dx = ∫ tan x(sec^{2} x) sec^{2} x dx
= ∫ tan x(1 + tan^{2} x) sec^{2} x dx
= ∫ tan x sec^{2} x dx + ∫ tan^{3} x sec^{2} x dx
= ∫ u du + ∫ u^{3} du  (for u = tan x)
= \frac{u^{2}}{2} + \frac{u^{4}}{4} + C
= \frac{tan^{2} x}{2} + \frac{tan^{4} x}{4} + C.

This is equivalent to the answer above via the trig identity 1 + tan^{2} x = sec^{2} x.

Method 3: A third way to do this one is to convert everything to sines and cosines. The integral obtained that way can be done by a u-substitution letting u = cos x, du = − sin x dx:

∫ \frac{sin x}{cos^{5} x} dx = − ∫ u^{−5} du = \frac{u^{−4}}{4} + C = \frac{1}{4 cos^{4} x} + C.

This is also equivalent to the answer in Method 1 above since sec x = \frac{1}{cos x}.

B. (10) ∫ \frac{\sqrt{9 + e^{2x}}}{e^{x}} dx

Solution: Substitute u = e^{x}, then du = e^{x}dx, so dx = \frac{du}{e^{x}} = \frac{du}{u}. Making these substitutions yields

∫ \frac{\sqrt{9 + u^{2}}}{u^{2}} du,

which is # 24 in the table with a = 3. We apply the table entry:

= −\frac{\sqrt{9 + u^{2}}}{u} + ln |u + \sqrt{9 + u^{2}}| + C = −\frac{\sqrt{9 + e^{2x}}}{e^{x}} + ln |e^{x} + \sqrt{9 + e^{2x}}| + C.

Note that this is not # 23 because of the “extra” u in the denominator from dx.

C. (10) ∫ \frac{dx}{x^{2} + 8x + 17}.

Solution: Complete the square in the denominator: x^{2} + 8x + 17 = (x + 4)^{2} + 1. Therefore, the substitution u = x + 4, du = dx takes this to the basic form

∫ \frac{1}{u^{2} + 1} du = tan^{−1}(u) + C = tan^{−1}(x + 4) + C.